# Solution to Legendre's Differential Equation

## Theorem

The solution of Legendre's differential equation:

$\displaystyle \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - 2 x \frac {\d y} {\d x} + p \paren {p + 1} y = 0$

can be obtained by Power Series Solution Method.

## Proof

Let:

$\displaystyle y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$

such that:

$a_0 \ne 0$
$\displaystyle \dot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1}$
$\displaystyle \ddot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$

Substituting in the original equation:

 $\displaystyle \paren {1 - x^2} \ddot y - 2 x \dot y + p \paren {p + 1} y$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {1 - x^2} \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - x^2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$  $\displaystyle$ $\displaystyle {} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}$ $=$ $\displaystyle 0$

The summations are dependent upon $n$ and not $x$.

Therefore it is a valid operation to multiply the $x$'s into the summations, thus:

 $\displaystyle \leadsto \ \$ $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$  $\displaystyle$ $\displaystyle {} - \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n} - 2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$  $\displaystyle$ $\displaystyle {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n - 1} - 2 \paren {k - n} }$ $=$ $\displaystyle 0$

Increasing the summation variable $n$ to $n + 2$:

 $\displaystyle \leadsto \ \$ $\displaystyle \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}$  $\displaystyle$ $\displaystyle {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }$ $=$ $\displaystyle 0$

Taking the first 2 terms of the second summation out:

 $\displaystyle \leadsto \ \$ $\displaystyle \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}$  $\displaystyle$ $\displaystyle {} + a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }$  $\displaystyle$ $\displaystyle {} + \sum_{n \mathop = 2}^\infty a_{n - 2} x^{k - n} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {\paren {p + 1} - k \paren {k - 1} }$  $\displaystyle$ $\displaystyle {} + \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }$ $=$ $\displaystyle 0$

Equating each term to $0$:

 $(1):\quad$ $\displaystyle a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} }$ $=$ $\displaystyle 0$ $(2):\quad$ $\displaystyle a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }$ $=$ $\displaystyle 0$ $(3):\quad$ $\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }$ $=$ $\displaystyle 0$

Take equation $(1)$:

$a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } = 0$

It is assumed that $a_0 \ne 0$ and $x^k$ can never be zero for any value of $k$.

Thus:

 $\displaystyle p \paren {p + 1} - k \paren {k + 1}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle p^2 - k^2 + p - k$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {p - k} \paren {p + k} + \paren {p - k}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {p - k} \paren {p + k + 1}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle p - k$ $=$ $\displaystyle 0$ $\, \displaystyle \lor \,$ $\displaystyle p + k + 1$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle k$ $=$ $\displaystyle p$ $\, \displaystyle \lor \,$ $\displaystyle k$ $=$ $\displaystyle p - 1$

Take equation $(2)$:

$a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} } = 0$

As before, it is assumed that $x^{k - 1}$ can never be zero for any value of $k$.

Thus:

 $\displaystyle a_1 \paren {p \paren {p + 1} - k \paren {k - 1} }$ $=$ $\displaystyle 0$ $\displaystyle a_1 k$ $=$ $\displaystyle 0$ substituting the value of $k$ from $(1)$ $\displaystyle a_1$ $=$ $\displaystyle 0$ as $2 k \ne 0$

Take equation $(3)$:

$\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } } = 0$

As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$.

Thus:

 $\displaystyle 0$ $=$ $\displaystyle a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }$ $\displaystyle \leadsto \ \$ $\displaystyle a_n$ $=$ $\displaystyle -\frac {\paren {k - n + 2} \paren {k - n + 1} } {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } a_{n - 2}$

Since Legendre's differential equation is a second order ODE, it has two independent solutions.

### Solution 1 (for $k = p$)

 $\displaystyle y$ $=$ $\displaystyle x^p \sum_{n \mathop = 0}^\infty a_n x^{-n}$ $\displaystyle$ $=$ $\displaystyle x^k \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}$

 $\displaystyle a_n$ $=$ $\displaystyle -\frac {\paren {p - n + 2} \paren {p - n + 1} } {p \paren {p + 1} - \paren {p - n} \paren {p - n + 1} } a_{n - 2}$ from equation $(4)$ $\displaystyle$ $=$ $\displaystyle -\frac {\paren {p - n + 2} \paren {p - n + 1} } {n \paren {2 p - n + 1} } a_{n - 2}$

Thus:

 $\displaystyle a_2$ $=$ $\displaystyle -\frac {p \paren {p - 1} } {2 \paren {2 p - 1} } a_0$ for $n = 2$ $\displaystyle a_4$ $=$ $\displaystyle -\frac {\paren {p - 2} \paren {p - 3} } {4 \left({2 p - 3}\right)} a_2$ for $n = 4$ $\displaystyle$ $=$ $\displaystyle \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \left({2 p - 3}\right)} a_0$ $\displaystyle a_6$ $=$ $\displaystyle -\frac {\paren {p - 4} \paren {p - 5} } {6 \left({2 p - 5}\right)} a_4$ for $n = 6$ $\displaystyle$ $=$ $\displaystyle -\frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \paren {p - 4} \paren {p - 5} } {2 \cdot 4 \cdot 6 \paren {2 p - 1} \left({2 p - 3}\right) \left({2 p - 5}\right)} a_0$

Similarly:

 $\displaystyle a_{2 n}$ $=$ $\displaystyle \left({-1}\right)^n \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \dotsm \left({p - 2 n - 1}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \paren {2 p - 1} \left({2 p - 3}\right) \left({2 p - 3}\right) \dotsm \left({2 p - 2 n + 1}\right)} a_0$

Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$

Substituting for $a_n$:

 $\displaystyle y$ $=$ $\displaystyle a_0 x^k \left({1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \left({2 p - 3}\right)} x^{-4} }\right.$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \dotsb$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left.{\left({-1}\right)^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \left({p - 2 n + 1}\right)} {\left({2 \cdot 4 \cdot 6 \dots 2 n}\right) \paren {2 p - 1} \left({2 p - 3}\right) \dotsm \left({p - 2 n + 1}\right)} x^{-2 n} + \dotsb}\right)$

$\Box$

### Solution 2 (for $k = p - 1$)

 $\displaystyle y$ $=$ $\displaystyle x^{-p - 1} \sum_{n \mathop = 0}^\infty a_n x^{-n}$ $\displaystyle$ $=$ $\displaystyle x^{-k - 1} \left({a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\right)$

 $\displaystyle a_n$ $=$ $\displaystyle -\frac {\left({-p - 1 - n + 2}\right) \left({-p - 1 - n + 1}\right)} {p \paren {p + 1} - \left({- p - 1 - n}\right) \left({-p - n}\right)} a_{n - 2}$ from equation $(4)$ $\displaystyle$ $=$ $\displaystyle \frac {\left({p + n - 1}\right) \left({p + n}\right)} {n \left({2 p + n + 1}\right)} a_{n - 2}$

Thus:

 $\displaystyle a_2$ $=$ $\displaystyle \frac {\paren {p + 1} \paren {p + 2} } {2 \left({2 p + 3}\right)} a_0$ for $n = 2$ $\displaystyle a_4$ $=$ $\displaystyle \frac {\paren {p + 3} \paren {p + 4} } {4 \left({2 p + 5}\right)} a_2$ for $n = 4$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \left({2 p + 3}\right) \left({2 p + 5}\right)} a_0$ $\displaystyle a_6$ $=$ $\displaystyle \frac {\paren {p + 5} \paren {p + 6} } {6 \left({2 p + 7}\right)} a_4$ for $n = 6$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} \paren {p + 5} \paren {p + 6} } {2 \cdot 4 \cdot 6 \left({2 p + 3}\right) \left({2 p + 5}\right) \left({2 p + 7}\right)} a_0$

Similarly:

 $\displaystyle a_{2 n}$ $=$ $\displaystyle \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \left({p + 2 n}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p + 3}\right) \left({2 p + 5}\right) \left({2 p + 7}\right) \dotsm \left({2 p + 2 n + 1}\right)} a_0$

Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$

Substituting for $a_n$:

 $\displaystyle y$ $=$ $\displaystyle a_0 x^{-k - 1} \left({1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \left({2 p + 3}\right)} x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \left({2 p + 3}\right) \left({2 p + 5}\right)} x^{-4} }\right.$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \dotsb$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left.{\left({-1}\right)^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \left({p + 2 n}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p + 3}\right) \left({2 p + 5}\right) \dotsm \left({p - 2 n}\right)} x^{-2 n} + \dotsb}\right)$

$\Box$

In summary, the two particular solutions are:

 $\displaystyle y$ $=$ $\displaystyle a_0 x^k \left({1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \left({2 p - 3}\right)} x^{-4} }\right.$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \dotsb$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left.{\left({-1}\right)^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \left({p - 2 n + 1}\right)} {\left({2 \cdot 4 \cdot 6 \dots 2 n}\right) \paren {2 p - 1} \left({2 p - 3}\right) \dotsm \left({p - 2 n + 1}\right)} x^{-2 n} + \dotsb}\right)$

and:

 $\displaystyle y$ $=$ $\displaystyle a_0 x^{-k - 1} \left({1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \left({2 p + 3}\right)} x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \left({2 p + 3}\right) \left({2 p + 5}\right)} x^{-4} }\right.$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \dotsb$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left.{\left({-1}\right)^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \left({p + 2 n}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p + 3}\right) \left({2 p + 5}\right) \dotsm \left({p - 2 n}\right)} x^{-2 n} + \dotsb}\right)$

$\blacksquare$

## Also see

The solutions of Legendre's differential equation are known as Legendre polynomials, which are functions of the parameter $p$.

## Source of Name

This entry was named for Adrien-Marie Legendre.